A square piece of tile has a width of 13 cm. What is the area of the piece of tile?

Mathematics

2 answers:

sesenic [268]3 years ago

6 0

The area of a square is the square of its width.
A = (13 cm)² = 169 cm²

inessss [21]3 years ago

5 0

Square= 4 sides with same measurement

So if the width is 13 cm, all sides are 13 cm.

AREA OF SQUARE
A= a^2
A= (13)^2
A= 169 cm^2

ANSWER: The area of the square tile is 169 cm^2.

Hope this helps! :)

You might be interested in

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

6 0

3 years ago

Is 5:6 And 45:54 is equivalent or not equivalent

Anon25 [30]

Yes they r equivalent

4 0

2 years ago

Read 2 more answers

8. Sam divided a rectangle into 8 congruent

S_A_V [24]

Sam divided a rectangle into 8 congruent rectangles that each have a area of 5 cm2. what is the area of the rectangle before it is divided?

Answer: